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densk [106]
2 years ago
7

Every object that has mass attracts every other object with a gravitational force: It has been proven that the size of the

Physics
2 answers:
sergejj [24]2 years ago
8 0
Answer;

- This statement about matter and its behavior is best classified as a Law.
-It is the Law of Universal Gravitation.

Explanation;
-The Law of Universal Gravitation states that every point mass attracts every other point mass in the universe by a force pointing in a straight line between the centers-of-mass of both points, and this force is proportional to the masses of the objects and inversely proportional to their separation.This attractive force always points inward, from one point to the other.
-The Law applies to all objects with masses, big or small. Two big objects can be considered as point-like masses, if the distance between them is very large compared to their sizes or if they are spherically symmetric.
denpristay [2]2 years ago
7 0

Answer:

this is a law because it is a constant fact of nature

Explanation:

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If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?
sdas [7]
Data:
f_{2} = 42 Hz
n (Wave node)
V (Wave belly) 
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
f_{n} =  \frac{nV}{2L}

Wire 2 → 2º Harmonic → n = 2

f_{n} = \frac{nV}{2L}
f_{2} = \frac{2V}{2L} &#10;
2V =  f_{2} *2L
V =  \frac{ f_{2}*2L }{2}
V =  \frac{42*2L}{2}
V =  \frac{84L}{2}
V = 42L

Wire 1 → 1º Harmonic or Fundamental rope → n = 1


f_{n} = \frac{nV}{2L}
f_{1} = \frac{1V}{2L}
f_{1} =  \frac{V}{2L}

If, We have:
V = 42L
Soon:
f_{1} = \frac{V}{2L}
f_{1} = \frac{42L}{2L}
\boxed{f_{1} = 21 Hz}

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

7 0
3 years ago
Read 2 more answers
What type of data allows geologists to know what rock layers lie underneath Minnesota?
mrs_skeptik [129]
The answer is well log data, it is a detailed log of information taken from a borehole which geologist used to study geological formations of the earth's layer taken from samples returned from the borehole which was dugged.
5 0
3 years ago
A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel
trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

1\ mile=1609.34\ m

1\ h=3600\ seconds

18.03\times \frac{3600}{1609.34}=40.33\ mph

Speed of the boat by when it will hit the dock is 40.33 mph

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m

The distance at which the boat will have to start decelerating is 312.5 m

5 0
2 years ago
The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the t
Nina [5.8K]

Answer:

The speed of the boxes are 1 m/s.

Explanation:

Given that,

Mass of box = 1 kg

Mass of another box = 2 kg

Suppose 1 kg box moves with 3 m/s speed.

We need to calculate the speed of the boxes

Using formula of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v

Where, u = initial velocity

v = final velocity

Put the value into the formula

1\times3+2\times0=(1+2)v

v=\dfrac{3}{3}

v=1\ m/s

Hence, The speed of the boxes are 1 m/s.

4 0
3 years ago
Calculate the power of a car that does 50,000 J of work in 7s during a race
Fantom [35]
Answer =7,142.9W

I hope this helps (:
6 0
3 years ago
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